5 THE MEAN VALUE THEOREM 159

Let n -> pn be a bijection of N onto D; for any g > 0, we will prove that
11/08) -/(«)!! < 9(P) ~ 9(a) + e(j8 - a + 2); the left hand side being inde-
pendent of 8, this will complete the proof, Define A as the subset of I consisting
of the points f such that, for a < f < £,

It is clear that a e A; if £ e A and a < ^ < {, then 77 e A also, by definition;
this shows that if y is the l.u.b. of A, then A must be either the interval [a, y[
or the interval [a, y]; but in fact, from the definition of A it follows at once
that A = [a, y]. Moreover, from the continuity of /and <p it follows that

(8.5.1 .1) ||/(7) -/(a)|| < <p(j) - <p(a) + e(y - a) + e £ 2~"

Pn<7

and therefore we need only prove that y = /?. Suppose y < ft; if y^D,
then from the definition of the derivative, it follows that there is an interval
[y, y + A] contained in I such that, for y < f < y + X

ll/(0 -/(y) -/W - y)ll < (fi/2)(f - y)
and

hence

II/CO -/(T)ll < II/'(?)II(C - 7) + (e/2)(C - y)

< fl>'(vXC - r) + («/2XC - ?) < 9(0 - 9(7) + «<C - 7)

and from (8.5.1.1) we deduce

ll/(0 -/(«)!! < 9(0 - 9(«) + e(C - «) + e Z 2-"

Pn<V

contrary to the definition of y.Ifye D, let y = pm; it follows from the con-
tinuity of /and (p that there is an interval [7, 7 -f I] contained in I, such that
for y ^ C < 7 + ^

hence, from (8.5.1.1) we deduce again

and we reach again a contradiction. Q.E.D.
The most important case is that in which cp(g) = M(£ — a) with M > 0: A = 0 and such that z0 =0; observe that if tj, is the largest number in I (or the